clock in the moving coordinate system reads t’ = 0 when the stationary clock reads t = 0. If the moving frame moves at a speed of 0.800c, what time will the moving clock read when the stationary observer reads 15.0 hr on her clock?
We know that if one frame is moving with a speed v = 0.800c, then the time between the two will vary by a factor of
"\\sqrt{(1-v^{2}\/c^{2})}= \\sqrt{(1-(0.800c)^{2}\/c^{2})}=\\sqrt{(1-0.800^{2})}=0.6"
Thus, the time measured in system accelerating will be 0.6 times of the time measured in system at rest.
So, the time measured in accelerating system is
"t^{'}=0.6\\times 15=9hr"
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